What is the integral from-5 to 5 of (sqrt(25-x^2)) ?

The integral from-5 to 5 of (sqrt(25-x^2)) is (25pi)/2

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integral from-5 to 5 of (sqrt(25-x^2))

\int_{- 5}^{5} (25 - x^{2}) d x

\frac{25 π}{2}

Decimal Notation

39.26990 \dots

Solution steps

\int_{- 5}^{5} 25 - x^{2} d x

Apply Trigonometric Substitution

: \int_{- \frac{π}{2}}^{\frac{π}{2}} 25 cos^{2} (u) d u

= \int_{- \frac{π}{2}}^{\frac{π}{2}} 25 cos^{2} (u) d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 25 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} cos^{2} (u) d u

Rewrite using trig identities

= 25 \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} \frac{1 + cos ( 2 u )}{2} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 25 \cdot \frac{1}{2} \cdot \int_{- \frac{π}{2}}^{\frac{π}{2}} 1 + cos (2 u) d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 25 \cdot \frac{1}{2} (\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u + \int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 u) d u)

\int_{- \frac{π}{2}}^{\frac{π}{2}} 1 d u = π

\int_{- \frac{π}{2}}^{\frac{π}{2}} cos (2 u) d u = 0

= 25 \cdot \frac{1}{2} (π + 0)

Simplify

25 \cdot \frac{1}{2} (π + 0) : \frac{25 π}{2}

= \frac{25 π}{2}

\int \frac{1}{( 4 x + 5 ) ^{10}} d x

y^{^{''}} + 2 y^{^{'}} + 10 y = 0

f (x) = arcsin (5 x)

x \to 5 lim (\frac{x - 5}{2 x - 10})

x \to - 1 lim (\frac{x ^{2} - 9}{x + 1})

What is the integral from-5 to 5 of (sqrt(25-x^2)) ?
The integral from-5 to 5 of (sqrt(25-x^2)) is (25pi)/2